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Protein pH titration (Henderson-Hasselbalch)

  1. Aug 18, 2009 #1
    1. The problem statement, all variables and given/known data

    A solution of histidine has all acid-base groups protonated. The pKa of the carboxy group is 1.8, the amino group is 9.2, and the side chain is 6.0. For each mole of amino acid, 2.5 moles of NaOH are added. What is the pH of the solution?

    2. Relevant equations

    Henderson-Hasselbalch equation: pH = pKa + log ([A-]/[HA])

    3. The attempt at a solution

    I am trying to answer this question for a practice exam. I know that the base will deprotonate the protein, starting from the carboxy group followed by the side chain. I assume that the NaOH is very basic, so it will definitely cause a pH of greater than 6.0. How do I proceed from there?

    My attempt: pH = unknown
    pKa = 1.8, 6.0, 9.2
    [A-]/[HA] = 2.5

    pH = 9.2 + log(2.5) = 9.59, but this is wrong.

    Could someone please help me with this problem. Thank you very much.
     
  2. jcsd
  3. Aug 18, 2009 #2

    Borek

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    Staff: Mentor

    Honestly - no idea how to solve the question. The most obvious approach will be to calculate initial pH and continue from there, just calculating stoichiometry and NaOH excess. However...

    This is not a quantitative statement. All groups are protonated - no matter wht pH is, it is just a matter of where the equilibrium lies. At pH 10 there is one caroboxylic group protonated for every 108 molecules, at pH 0.8 there is one NOT protonated carboxylic group for evey nine protonated.

    Could be they want us to assume molecule like H3Histidine3+ that reacts with excess base. That's wrong, as such protonated molecule will exist only in the presence of huge excess of strong acid, which can't be ignored when calculating final concentration of the base.

    So - either I have not yet waken up, or the question has no solution as worded.
     
    Last edited by a moderator: Aug 13, 2013
  4. Aug 18, 2009 #3
    According to my professor, adding alkali neutralizes protons so that...
    Adding 0.5 mol of alkali will give a pH of the carboxyl group pKa = 1.8
    Adding 1.0 mol of alkali will give a pH between 1.8 and 6.0
    Adding 1.5 mol of alkali will give a pH of the histidine group side chain pKa = 6.0
    Adding 2.0 mol of alkali will give a pH between 6.0 and 9.2
    Adding 2.5 mol of alkali will give a pH of the amino group pKa = 9.2
    Adding 3 mol of alkali will fully deprotonate the histidine.

    I'm confused as to how he go the 0.5 mol increments of alkali.
     
  5. Aug 18, 2009 #4

    Borek

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    Staff: Mentor

    Take a look at the Henderson-Hasselbalch equation - at pH=pKa acid is exactly 50% neutralized ([HA]=[A-]).

    Trick is - while he is right about neutralization stoichiometry, he still has no idea what the final pH is.
     
    Last edited by a moderator: Aug 13, 2013
  6. Aug 18, 2009 #5

    chemisttree

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    Science Advisor
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    Gold Member

    Treat the hisitidine exactly as you would treat a phosphoric acid buffer problem (H3PO4). Histidine can be though of as a triprotic acid in this example.

    Have you done any problems using phosphoric acid in a like manner? For example, what would the pH be for a phosphoric acid solution that has had 2.5 equivalents of NaOH added? The pKa's are different but not the approach.

    Borek, I know you know how to do this problem!
     
  7. Aug 18, 2009 #6

    Borek

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    Staff: Mentor

    What I don't like about the question, is the wording.

    Would you say that phosphoric acid in its solution is fully protonated? 0.1M solution has pH around 1.6 - that means around 25% is in the form of H2PO4-. Perhaps that's nitpicking, but I still don't like it.
     
    Last edited by a moderator: Aug 13, 2013
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